OBJECTIVES

When you finish this section, you should be able to:

1. Use Comparison Tests for Convergence and Divergence (p. 576)
2. Use the Limit Comparison Test (p. 577)

RECALL

The General Convergence Test: An infinite series of positive terms converges if and only if its sequence of partial sums is bounded.

THEOREM Comparison Test for Convergence

If \(0<a_{k}\leq b_{k}\) for all \(k\) and \(\sum\limits_{k=1}^{\infty}b_{k}\) converges, then \(\sum\limits_{k=1}^{\infty}a_{k}\) converges.

Using the Comparison Test for Convergence

Show that the series below converges: \[ \sum\limits_{k=1}^{\infty }\frac{1}{k^{k}}=1+\frac{1}{2^{2}}+\frac{1}{3^{3}} +\cdots +\frac{1}{n^{n}}+\cdots \]

Solution We know that the geometric series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{2^{k}}\) converges since \(\vert r\vert =\dfrac{1}{2}<1\). Now, since \(\dfrac{1}{n^{n}}\leq \dfrac{1}{2^{n}}\) for all \(n\geq 2\), except for the first term, each term of the series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}\) is less than or equal to the corresponding term of the convergent geometric series. So, by the Comparison Test for Convergence, the series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}\) converges.

NOTE

Do you see why \(\dfrac{1}{n^{n}}\leq \dfrac{1}{2^{n}}\) when \(n \geq 2\)? \(n\geq 2\Rightarrow {n}^{n}\geq {2}^{n}\Rightarrow\) \(\dfrac{{ 1}}{{n}^{n}}\leq \dfrac{1}{{2}^{n}}\).

NOW WORK

Problem 5.

THEOREM Comparison Test for Divergence

If \(0<c_{k}\leq a_{k}\) for all \(k\) and \(\sum\limits_{k=1}^{\infty}c_{k}\) diverges, then \(\sum\limits_{k=1}^{\infty }a_{k}\) diverges.

Using the Comparison Test for Divergence

Show that the series \(\sum\limits_{k=1}^{\infty }\dfrac{k+3}{k(k+2)}\) diverges.

Solution Since \(\dfrac{k+3}{k(k+2)}\) has a factor \(\dfrac{1}{k}\), we choose to compare the given series to the harmonic series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k}\), which diverges. \[ \frac{n+3}{n(n+2)}=\left( {\frac{n+3}{n+2}}\right) \left( {\frac{1}{n}} \right) >\frac{1}{n} \]

It follows from the Comparison Test for Divergence that \(\sum\limits_{k=1}^{\infty }\dfrac{k+3}{k(k+2) }\) diverges.

NOW WORK

Problem 9.

THEOREM Limit Comparison Test

Suppose \(\sum\limits_{k=1}^{\infty }\,a_{k}\) and \(\sum\limits_{k=1}^{\infty }\,b_{k}\) are both series of positive terms.

If \(\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=L\), \(0<\) \(L<\infty ,\) then both series converge or both series diverge.

The Limit Comparison Test gives no guidance if \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}\) \(=0\) or \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}=\infty\) or \(\lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}\) does not exist.

Proof

If \(\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=L\,{>}\,0\), then for any \(\varepsilon\,{>}\,0,\) there is a positive real number \(N\) so that \(\left\vert \dfrac{a_{n}}{b_{n}}-L\right\vert <\varepsilon ,\) for all \(n>N.\) If we choose \(\varepsilon =\dfrac{L}{2},\) then \[ \begin{eqnarray*} \begin{array}{rl@{\qquad\hspace{18.2pt}}l} \left\vert \dfrac{a_{n}}{b_{n}}-L\right\vert &< \dfrac{L}{2} & \hbox{for all }n>N \\[13pt] \dfrac{L}{2}<\dfrac{a_{n}}{b_{n}} &< \dfrac{3L}{2} & \hbox{for all }n>N\, \end{array} \end{eqnarray*} \]

Since \(b_{n}>0\), \[ \begin{array}{rl@{\qquad}l} \dfrac{L}{2}b_{n}<a_{n} & <\dfrac{3L}{2}b_{n} & \hbox{for all }n>N \end{array} \]

If \(\sum\limits_{k=1}^{\infty }a_{k}\) converges, by the Comparison Test for Convergence, the series \(\sum\limits_{k=1}^{\infty }\left(\dfrac{L}{2}b_{k}\right)\) \(=\dfrac{L}{2}\sum\limits_{k=1}^{\infty }b_{k}\) also converges. It follows that the series \(\sum\limits_{k=1}^{\infty }b_{k}\) is also convergent.

If \(\sum\limits_{k=1}^{\infty }b_{k}\) converges, then so does \(\sum\limits_{k=1}^{\infty }\left(\dfrac{3L}{2}b_{k}\right).\) Since \(a_{n}<\dfrac{3L}{2}b_{n},\) for all \(n>N,\) by the Comparison Test for Convergence, \(\sum\limits_{k=1}^{\infty }a_{k}\) also converges.

Therefore, \(\sum\limits_{k=1}^{\infty }a_{k}\) and \(\sum\limits_{k=1}^{\infty}b_{k}\) converge together. Since one cannot converge and the other diverge, they must diverge together.

RECALL

If \(c\) is a nonzero real number and if \(\sum\limits_{k\,=\,1}^{\infty}{a}_{k}={S}\) is a convergent series, then the series \(\sum\limits_{k\,=\,1}^{\infty} ({ca}_{k})\) also converges. Furthermore, \(\sum\limits_{k\,=\,1}^{\infty}({ca}_{k})={c}\sum\limits_{k\,=\,1}^{\infty }{a}_{k}={cS}\).

Using the Limit Comparison Test

Determine whether the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}\) converges or diverges.

Solution We choose an appropriate \(p\)-series to use for comparison by examining the behavior of the series for large values of \(n\): \[ \begin{eqnarray*} \dfrac{1}{2n^{3/2}+5}=\dfrac{1}{n^{3/2}\left( 2+\dfrac{5}{n^{3/2}}\right) }= \dfrac{1}{n^{3/2}}\left( \dfrac{1}{2+\dfrac{5}{n^{3/2}}}\right) & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow }}}} }{\approx} & \dfrac{1}{n^{3/2}}\left(\dfrac{1}{2}\right)\\ \end{eqnarray*} \]

This leads us to choose the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}},\) which converges, and use the Limit Comparison Test with \[ a_{n}=\dfrac{1}{2n^{3/2}+5} \qquad \hbox{and}\qquad b_{n}=\dfrac{1}{n^{3/2}} \] \[ \lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}=\lim_{n\,\rightarrow \,\infty }\frac{\dfrac{1}{2n^{3/2}+5}}{\dfrac{1}{n^{3/2}}} =\lim\limits_{n\rightarrow \infty }\frac{n^{3/2}}{2n^{3/2}+5} =\lim\limits_{n\rightarrow \infty }\dfrac{1}{2+\dfrac{5}{n^{3/2}}}=\frac{1}{2} \]

Since the limit is a positive number and the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}}\) converges, then by the Limit Comparison Test, \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}\) also converges.

NOW WORK

Problem 15.

Using the Limit Comparison Test

Determine whether the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k}+2}{\sqrt{k^{3}+3k^{2}+1}}\) converges or diverges.

Solution We choose a \(p\)-series for comparison by examining how the terms of the series behave for large values of \(n\): \[ \begin{eqnarray*} \frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}=\frac{\sqrt{n}\,\left( 3+\dfrac{2}{ \sqrt{n}}\right) }{\sqrt{n^{3}}\left( \sqrt{1+\dfrac{3}{n}+\dfrac{1}{n^{3}}} \right) }=\dfrac{n^{1/2}}{n^{3/2}}\frac{3+\dfrac{2}{\sqrt{n}}}{\sqrt{1+ \dfrac{3}{n}+\dfrac{1}{n^{3}}}} & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow}}}}}{\approx} & \dfrac{1}{n}(3) \\ \end{eqnarray*} \]

So, we compare the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k~}+2}{\sqrt{k^{3}+3k^{2}+1}}\) to the harmonic series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\), which diverges, and use the Limit Comparison Test with \[ a_{n}=\frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}\qquad \hbox{ and }\qquad b_{n}=\dfrac{1}{n} \] \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}} &=& \lim_{n\,\rightarrow \,\infty }\frac{\dfrac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}}{\dfrac{1}{n}} =\lim\limits_{n\rightarrow \infty }\frac{n\left( 3\sqrt{n}+2\right) }{\sqrt{ n^{3}+3n^{2}+1}}=\lim\limits_{n\rightarrow \infty }\frac{3n^{3/2}+2n}{\sqrt{n^{3}+3n^{2}+1}}\\[7pt] &=&\lim\limits_{n\rightarrow \infty }\dfrac{3n^{3/2}}{n^{3/2}}=3 \end{eqnarray*} \]

Since the limit is a positive real number and the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\) diverges, then by the Limit Comparison Test, \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k~}+2}{\sqrt{k^{3}+3k^{2}+1}}\) also diverges.

NOW WORK

Problem 17.